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QUESTIONS AND ANSWERS
CHAPTER 6
1.
Define angle modulation
Angle modulation results whenever the phase angle of a sinusoidal wave is varied with
respect to time.
2.
Define direct FM and indirect FM.
Direct FM results whenever the frequency of a constant amplitude carrier is varied
directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of
the modulating signal. Indirect FM is a direct PM.
3.
Define direct PM and indirect PM.
Direct PM results whenever the phase of a constant amplitude carrier is varied directly
proportional to the amplitude of the modulating signal at a rate equal to the frequency of the
modulating signal. Indirect PM is an direct FM.
4.
Define frequency deviation and phase deviation.
Frequency deviation is the relative displacement of the carrier frequency in hertz in
respect to its unmodulated value. Phase deviation is the relative angular displacement of the
carrier phase in radians in respect to the reference phase.
5.
Define instantaneous phase, instantaneous phase deviation, instantaneous frequency, and
instantaneous frequency deviation,
Instantaneous phase deviation is the instantaneous change in the phase of the carrier at
a given instant of time and indicates how much the phase of the carrier is changing with respect
to its reference phase. Instantaneous phase is the precise phase of the carrier at a given instant
of time. Instantaneous frequency deviation is the instantaneous change in the frequency of the
carrier and is defined as the first time derivative of the instantaneous phase deviation.
Instantaneous frequency is the precise frequency of the carrier at a given instant of time.
6.
Define deviation sensitivity for a frequency modulator and for a phase modulator.
Deviation sensitivity is the output-versus-input transfer functions for the modulators,
which give relationship between what output parameter changes in respect to specified changes
in the input signal.
For FM, changes would occur in the output frequency in respect to changes in
amplitude to the input voltage.
For PM, changes would occur in the phase of the output frequency in respect to
changes in the amplitude of the input voltage.
7.
Describe the relationship between the instantaneous carrier frequency and the modulating signal
for FM.
The instantaneous carrier frequency is directly proportional to the amplitude of the
modulating signal.
8.
Describe the relationship between the instantaneous carrier frequency and the modulating signal for FM.
The instantaneous carrier frequency is directly proportional to the amplitude of the
modulating signal.
9.
Describe the relationship between frequency deviation and the amplitude and frequency of the
modulating signal.
Frequency deviation is directly proportional to the amplitude of the modulating signal and
inversely proportional to its frequency.
10.
Define carrier swing.
Carier swing is the peak to peak carrier swing.
11.
Define modulation index for FM and for PM.
For FM, modulation index is the ratio between the frequency deviation and the frequency
of the modulating signa. For PM, modulation index is the product of the modulating signal voltage
and the deviation sensitivity.
12.
Describe the relationship between modulation index and the modulating signal for FM; for PM.
For FM, modulation index is in direct proportion to the amplitude of the modulating signal
and inverse proportion to the frequency. For PM, modulation index is in direct proportion to the
amplitude of the modulating signal, independent of its frequency.
13.
Define percent modulation for angle- modulated signals.
Percent modulation for angle modulation is the ratio of the frequency actually produced to
the maximum frequency deviation allowed by law.
14.
Describe the difference between a direct frequency modulator and a direct phase modulator.
Direct frequency modulator is a circuit in which the carrier is varied in such a way that
instantaneous phase is proportional to the integral of the modulating signal. Direct phase
modulator is a circuit in which the carrier is varied in such away that instantaneous phase is
proportional to the modulating signal.
15.
How can a frequency modulator be converted to a phase modulator; a phase modulator to a
frequency modulator?
FM modulator = integrator followed by a PM modulator. PM modulator = differentiator
followed by a FM modulator.
16.
How many sets of sidebands are produced when a carrier is frequency modulated by a single
input frequency?
There are infinite number of pairs of side frequency are produced, therefore infinite
bandwidth.
17.
What are the requirements for a side frequency to be considered significant?
To be considered significant, a side frequency must have an amplitude equal to or
greater than one percent of the unmodulated carrier amplitude.
18.
Define a low, a medium, a high modulation index.
In low index case, modulation index is less than 1.Most of the signal information is carried
by the first set of sidebands. For high index case, modulation index is greater than 10. the
minimum bandwidth required to propagate the frequency modulated wave is equal to peak to
peak frequency deviation. While modulation index greater than 1 but less than 10 are classified
as medium index.
19.
Describe the significant of the Bessel table.
Bessel table defines the actual minimum bandwidth required to pass all the significant
sideband sets. It shows several values of modulation index and the corresponding sets of side
frequencies.
20.
State Carson’s general rule for determining the bandwidth foer an angle- modulated wave.
Carson’s rule approximate the bandwidth necessary to transmit an angle modulated
waves as twice the sum of the peak frequency deviation and the highest modulating signal
frequency.
21.
Define deviation ratio.
Deviation ratio is the worst case modulation index and is equal to the maximum peak
frequency deviation divided by the maximum modulating signal frequency
22.
Describe the relationship between the power in the unmodulated carrier and the power in the
modulated wave for FM.
Unlike AM, the total power in an unmodulated wave for FM is equal to the power of
the unmodulated carrier.
23.
Describe the significance of the FM noise triangle.
The FM noise triangle is significant because it shows the noise voltage at the output of a
PM demodulator is constant with frequency, whereas, the noise voltage at the output of an FM
demodulator increases linearly with frequency.
24.
What effect does limiting have on the composite FM waveform?
The limiters are very effective in removing the amplitude variations cause by noise from
the FM composite waveform by clipping the peaks of the envelope prior to detection.
25.
Define preemphasis and deemphasis.
With FM, there is a non-uniform distribution of noise. To compensate for this
preemphasis and deemphasis is used. Preemphasis allows the high frequency modulating signal
to modulate the carrier at a highre level and thus causes more frequency deviation than their
original amplitudes. While deemphasis restores the original amplitude versus frequency
characteristics to the information signals.
26.
Describe a preemphasis network; deemphasis network.
Preemphasis network is a high pass filter called a differentiator. It provides a constant
increase in the amplitude of the modulating signal with an increase in frequency. Deemphasis
network is a low pass filter called an integrator.
27.
Describe the basic operation of a varactor diode FM generator.
The operation of a varactor diode FM generator goes this way: R1 & R2, develop a dc
voltage that reverse biases the varactor diode and determines the rest frequency of the oscillator.
The external modulating signal voltages to and subtracts from the dc bias, which changes the
capacitance and increases the frequency of oscillation. Positive alternations of the modulating
signal increase the reverse bias on the varactor, which decreases its capacitance and increases
the frequency of oscillation. Conversely, negative alternations decrease the frequency of
oscillation.
28.
Describe the basic operation of a reactance FM modulator.
The modulating signal varies the reactance of Q 1, which causes a corresponding change
in the resonant frequency of the oscillator tank circuit. When a modulating signal is applied to
bottom of R3, the gate-to-source voltage is varied accordingly, causing proportional change in gm.
As a result, equivalent circuit impedance is a function of the modulating signal.
29.
Describe the basic operation of a linear integrated- circuit FM modulator.
The operation of a linear integrated circiut FM modulator goes this way: The voltage
controlled oscilator frequency is determined by external resistor and capacitor (R & C). The input
modulating signal is applied directly to the input of the
30.
Draw the block diagram for a crosby direct FM transmitter and describe its operation.
Frequency modulator
Modulating
signal input
Frequency modulator
and master oscillator
N1
N3
N2
Power
amplifier
to antenna
LPF
Discriminator
BPF
Mixer
AFC loop
Crystal
oscillator
Buffer
CROSBY DIRECT FM TRANSMITTER
31.
What is the purpose of AFC loop? Why is one required for the crosby transmitter?
The purpose of AFC loop is to achieve a near crystal stability of the transmit carrier
frequency without using a crystal in the carrier oscillator. It is required for Crosby transmitter
because this transmitter uses either VCO, a reactance oscillator, or a linear IC to generate the
carrier frequency, thus, it is more susceptible to frequency drift due to temperature change, power
supply fluctuations and so on. But although this circuit does not totally eliminate frequency drift, it
can substantially reduce it.
32.
Draw the block diagram for a phase- locked- loop Fm transmitter and describe its operation.
fo
Nfo
Divide by N
Crystal
Reference
FM output
Phase comparator
Oscillator
VCO
Phase-locked loop
DC correction
voltage
Low-pass
dc + ac
Summer
filter
Modulating
signal input
PHASE-LOCKED-LOOP FM TRANSMITTER
The VCO output frequency is divided by N and fed back to the PLL phase comparator
where it is compared to a stable crystal reference frequency. The phase comparator
generates a correction voltage that is proportional to the difference between two frequencies.
The correction voltage adjusts the VCO center frequency to its proper value. Again, the LPF
prevents changes in the VCO output frequency due to the modulating signal from being
converted to a voltage, fed back to the VCO, and wiping out the modulation. The LPF also
prevents the loop from locking into a side frequency.
33.
Draw the block diagram for an Armstrong indirect FM transmitter and describe its operation.
ft
Power
Amplifier
Bandpass
filter
ft
X 72
Multiplier
Crystal
carrier
oscillator
Bandpass
filter
Vc
Buffer
amplifier
f1
Combining
network
Bandpas
s filter
X 72
Multiplier
Vusf + Vlsf = Vm
90
phase
shifter
V’c
Balanced
modulator
f2
Mixer &
downconverter
fo
Buffer
amplifier
fo
Modulating
signal input
Crystal
oscillator
ARMSTRONG INDIRECT FM TRANSMITTER
With an Armstrong (indirect) transmitter, a relatively low-frequency subcarrier (fc) is
phase shifted 90° (fc’) and fed to a balanced modulator, where it is mixed with the input
modulating signal (fm). The output from the balanced modulator is a double-sideband,
suppressed carrier wave that is combined with the original carrier in a combining network to
produce a low-index, phase-modulated waveform.
34.
Compare FM to PM.
The modulation index for FM is directly proportional to the amolitude of the modulating
signal and inversely proportional to its frequency while in PM, modulation index is directly
proportionalto the amplitude of modulating signal and independent of its frequency.
PROBLEMS
1. If a frequency modulator produces 5 kHz of frequency deviation for a 10-V modulating signal,
determine the deviation sensitivity. How much frequency deviation is produced for a 2-V
modulating signal?
If Em = 2 V
K = ∆f / Em
∆f = ?
= 5 kHz / 10 V
= K Em
K = 0.5 kHz/V
= 0.5 kHz/V (2V)
∆f = 1 kHz
2. If a phase modulator produces 2 rad of phase deviation for a 5-V modulating signal, determine
the deviation sensitivity. How much phase deviation would a 2-V modulating signal produce?
K
K
= ∆φ / Em
= 2 rad / 5 V
= 0.4 rad/V
If Em = 2 V
∆φ = ?
= K Em
= 0.4 rad/V (2V)
∆φ = 0.8 rad
3. Determine (a) the peak frequency deviation, (b) the carrier swing, and (c) the modulation index
for an FM modulator with deviation sensitivity K1 = 4 kHz/V and a modulating signal vm(t) = 10
sin(2π2000t). What is the peak frequency deviation produced if the modulating signal were to
double in amplitude?
a)
b)
∆f = KEm
= 4 kHz/V (10V)
∆f = 40 kHz
c)
2∆f = 2 (40 kHz)
= 80 kHz
FOR twice amplitude:
∆f
= 4 kHz (20 V)
∆f
= 80 kHz
m
m
= ∆f / fm
= 40 / 2
= 20
4. Determine the peak phase deviation for a PM modulator with a deviation sensitivity K = 1.5 rad/V
and a modulating signal vm(t) = 2 sin(2π2000t). How much phase deviation is produced for a
modulating signal with twice the amplitude?
FOR twice amplitude:
∆φ = 1.5 rad/V (2V)
∆φ = 1.5 rad (4V)
∆φ = 3 rad
∆φ = 6 rad
5. Determine the percent modulation for a television broadcast station with a maximum frequency
deviation ∆f = 50 kHz when the modulating signal produces 40 kHz of frequency deviation at the
antenna. How much deviation is required to reach 100% modulation of the carrier?
∆fmax = 10 kHz
∆factual = 2 kHz
% mod’n
= ∆factual /∆fmax x 100
= 40 kHz/ 50kHz x 100
= 80 %
FOR 100% modulation:
∆factual = 100 (50 kHz) / 100
= 50 kHz
6. From the Bessel table, determine the number of sets of sidebands produced for the following
modulation indices: 0.5,1,0,2.0.5.0, and 10.0.
Modulation index
0.5
1.0
2.0
5.0
10.0
# of sideband sets
2
3
4
8
14
7. For an FM modulator with modulation index m = 2, modulating signal vm(t) = Vm sin(2π2000t),
and an unmodulated carrier vc(t) = 8 sin(2π800kt),
(a) Determine the number of sets of significant sidebands.
(b) Determine their amplitudes.
(c) Draw the frequency spectrum showing the relative amplify
(d) Determine the bandwidth.
(e) Determine the bandwidth if the amplitude of the modulating signal increases by a factor
of 2.5.
m
Ec
fc
fm
a)
=2
=8V
= 800 kHz
= 2 kHz
b)
Amplitudes:
Jo = 8 (0.22)
J1 = 8 (0.58)
J2 = 8 (0.35)
J3 = 8 (0.13)
J4 = 8 (0.03)
From Bessel Table:
n = 4 sidebands
4.64
= 1.76
= 4.64
= 2.80
= 1.04
= 0.24
V
V
V
V
V
4.64
c)
2.80
1.76
2.80
1.04
1.04
0.24
792
d)
0.24
794
796
B
= 2nfm
= 2 (4) (2 kHz)
B
= 16 kHz
798
800
e)
802
804
806
kHz
808
FOR amplitude inc. by factor of 2.5:
m = 2 (2.5) = 5 .·. n = 8
B = 2 (8) (2 kHz)
B = 32 kHz
8. For an FM transmitter with 60-kHz carrier swing, determine the frequency deviation. If the
amplitude of the modulating signal decreases by a factor of 2, determine the new frequency
deviation.
carrier swing
2∆f = 60 kHz
∆f
= 60 kHz / 2
= 30 kHz
FOR amplitude decreased
by a factor of 2:
∆f
= K(Em/2)
= 30 kHz / 2
∆f
= 15 kHz
9. For a given input signal, an FM broadcast-band transmitter has a frequency deviation of ∆f = 20
kHz. Determine the frequency deviation if the amplitude of the modulating signal increases by a
factor of 2.5.
FOR amplitude decreased
by a factor of 2.5:
∆f
= K(Em x 2.5)
= 20 kHz (2.5)
∆f
= 50 kHz
∆f = KEm
10. An FM transmitter has a rest frequency fc = 96 MHz and a deviation sensitivity of K1 = 4 kHz/V.
Determine the frequency deviation for a modulating signal vm(t) = 8 sin(2π2000t). Determine the
modulation index.
fc
K
Em
fm
∆f = KEm
= 4 kHz/V (8V)
∆f = 32 kHz
= 96 MHz
= 4 kHz/V
=8V
= 2 kHz
m = ∆f / fm
= 32 / 2
m = 16
11. Determine the deviation ratio and worst-case bandwidth for an FM signal with a maximum
frequency deviation ∆f = 25 kHz and a maximum modulating signal fm(max) = 12.5 kHz.
∆fmax = 20 kHz
fmax = 12.5 kHz
B
B
= 2 (∆fmax + fmmzx)
= 2 (25 + 12.5)
= 75 kHz
DR = ∆fmax / fmmzx
= (25 / 12.5)
DR = 2
12. For an FM modulator with 40-kHz frequency deviation and a modulating-signal frequency fm= 10
kHz, determine the bandwidth using both the Bessel table and Carson’s rule.
∆f = 40 kHz
fm = 10 kHz
Using Bessel:
B
m = ∆f / fm
= 40 / 10
m =4
therefore,
n =7
= 2nfm
= 2 (7) (10 kHz)
= 75 kHz
B
Using Carson’s:
B = 2 (∆f + fm)
= 2 (40 + 10)
B = 100 kHz
13. For an FM modulator with an unmodulated carrier amplitude Vc = 20 V, a modulation index m =
1, and a load resistance RL = 10, determine the power in the modulated carrier and each side
frequency, and sketch the power spectrum for the modulated wave.
m =1
Amplitudes:
Jo
= 10 (0.77)
J1
= 10 (0.44)
J2
= 10 (0.11)
J3
= 10 (0.02)
= 15.4
= 8.80
= 2.20
= 0.04
V
V
V
V
Power:
Po
P1
P2
P3
= (15.4)2 / 20
= (8.80)2 / 20
= (2.20)2 / 20
= (0.04)2 / 20
= 11.858 W
= 3.8720 W
= 0.2420 W
= 0.0080 W
Power Spectrum:
11.858 W
3.872 W
3.872 W
0.242 W
0.242 W
0.008 W
0.008 W
14. For an angle-modulated carrier vc(t) = 2 cos(2π200MHz t) with 50 kHz of frequency deviation due
to the modulating signal and a single-frequency interfering signal Vn(t) = 0.5 cos(2π200.01 MHz
t), determine
(a) Frequency of the demodulated interference signal.
(b) Peak phase and frequency deviation due to the interfering signal.
(c) Signal-to-noise ratio at the output of the demodulator.
Ec
fc
∆f
En
fn
a)
c)
=2V
= 200 MHz
= 50 kHz
= 0.5 V
= 200.01 MHz
= fn – fc
= 200.01– 200
f
= 10 kHz
S/N due to interfering
tone:
S/N = Ec / En
= 2 / 0.5
=4
S/N after demodulation:
S/N = ∆fsig /∆fnoise
= 50 / 2.5
= 20
f
b)
∆fpeak = f (Vn / Vc)
= 10 kHz (0.5 / 2)
∆fpeak
= 2.5 kHz
∆φpeak
= Vn / Vc
= 0.5 / 2
∆φpeak = 0.25 rad
Voltage S/N improvement:
20 / 5 = 4
dB = 20 log 4
= 14 dB
15. Determine the total peak phase deviation produced by a 5-kHz band of random noise with a peak
voltage Vn = 0.08 V and a carrier vc(t) = 1.5 sin(2π40 MHz t).
Vn = 0.08 V
Vc = 1.5 V
∆φ = Vn / Vc
= 0.08 / 1.5
= 0.0533
rms:
= 0.0533 / √2
∆φ = 0.0377
18. If a frequency modulator produces 4 kHz of frequency deviation for a 10-Vp modulating signal,
determine the deviation sensitivity.
∆f = 4 kHz
Em = 10 Vp
K
K
= ∆f / Em
= 4 kHz / 10 V
= 400 Hz/V
19. If a phase modulator produces 1,5 rad of phase deviation for a 5-Vp modulating signal, determine the deviation sensitivity.
∆φ = 1.5 rad
Em = 5 Vp
K
K
= ∆φ / Em
= 1.5 rad / 5 V
= 0.3 rad/V
20. Determine (a) the peak frequency deviation, (b) the carrier swing, and (c) the modulation index
for an FM modulator with a deviation sensitivity K1 = 3 kHz/V and a modulating signal vm = 6
sin(2π2000t).
Em = 6 V
K = 3 kHz/V
fm = 2 kHz
a)
b)
2∆f = 2 (18 kHz)
= 36 kHz
c)
m = ∆f / fm
= 18 / 2
m =9
∆f = KEm
= 3 kHz/V (6V)
∆f = 18 kHz
21. Determine the peak phase deviation for a PM modulator with deviation sensitivity K = 2 rad/V
and a modulating signal vm = 4 sin(2π1000t).
∆φ = KEm
= 2 rad/V (4V)
∆φ = 8 rad
Em = 4 V
K = 2 rad/V
fm = 1 rad
22. Determine the percent modulation for a television broadcast station with a maximum frequency
deviation ∆f = 50 kHz when the modulating signal produces 30 kHz of frequency deviation.
∆fmax = 50 kHz
∆factual = 30 kHz
% mod’n
= ∆factual /∆fmax x 100
= 30 kHz/ 50kHz x 100
= 60
23. From the Bessel table determine the number of side frequencies produced for the following
modulation indices: 0.25,0.5,1.0, 2.0, 5.0, and 10.
Modulation index
0.25
0.5
1.0
2.0
5.0
10.0
# of sideband sets
1
2
3
4
8
14
24. For an FM modulator with modulation index m = 5, modulating signal vm == 2 sin(2π5kt), and an
unmodulated carrier frequency fc = 400 kHz, determine
(a) Number of sets of significant sidebands.
(b) Sideband amplitudes.
Then (c) Draw the output frequency spectrum.
m
Ec
fc
fm
=5
=2V
= 400 kHz
= 5 kHz
a)
From Bessel Table:
n = 8 sidebands
b)
Amplitudes:
Jo = 2 (-0.18)
J1 = 2 (-0.33)
J2 = 2 (0.50)
J3 = 2 (0.36)
= -0.36
= -0.66
= 1.00
= 0.72
J4
J5
J6
J7
J8
V
V
V
V
1.0
c)
= 2 (0.39)
= 2 (0.26)
= 2 (0.13)
= 2 (0.05)
= 2 (0.02)
V
V
V
V
V
1.0
0.78
0.52
= 0.78
= 0.52
= 0.26
= 0.10
= 0.04
0.78
0.72
0.72
0.52
0.1 0.26
0.04
0.26 0.1
-0.66 -0.36 -0.66
360 365 370 375 380 385 390
395 400
405 410 415
0.04
(kHz)
420 425 430 435 440
25. For an FM transmitter with an 80-kHz carrier swing, determine the frequency deviation. If the
amplitude of the modulating signal decreases by a factor of 4, determine the new frequency
deviation.
carrier swing
2∆f = 80 kHz
∆f
= 80 kHz / 2
= 40 kHz
FOR amplitude decreased
by a factor of 4:
∆f
= K(Em/4)
= 30 kHz / 4
∆f
= 10 kHz
26. For a given input signal, an FM broadcast transmitter has a frequency deviation ∆f = 40 kHz.
Determine the frequency deviation if the amplitude of the modulating signal increases by a factor
of 4.3.
FOR amplitude increased by a factor of 4.3:
∆f
= K(Em x 4.3)
= 40 kHz x 4.3
∆f
= 172 kHz
27. An FM transmitter has a rest frequency fc = 94 MHz and a deviation sensitivity K1 = 5kHz/V.
Determine the frequency deviation for a modulating signal vm(t) = 4 Vp.
Em = 4 Vp
fc = 94 MHz
∆f = KEm
= 5 kHz/V (4 V)
∆f = 172 kHz
28. Determine the deviation ratio and worst-case bandwidth for an FM system with a maximum
frequency deviation of 40 kHz and a maximum modulating-signal frequency fm = 10 kHz.
∆fmax = 40 kHz
fmmax = 10 kHz
DR = ∆fmax / fmmzx
= (40 / 10)
DR = 4
29. For an FM modulator with 50 kHz of frequency deviation arid a modulating-signal frequency fm =
8 kHz, determine the bandwidth using both the Bessel table and Carson's rule.
∆f = 50 kHz
fm = 8 kHz
Using Bessel:
B
m = ∆f / fm
= 50 / 8
m = 6.25 ≈ 6
therefore,
n =9
= 2nfm
= 2 (9) (8 kHz)
= 144 kHz
B
Using Carson’s:
B = 2 (∆f + fm)
= 2 (50 + 8)
B = 116 kHz
30. For an FM modulator with an unmodulated carrier voltage vc = 12 Vp, a modulation index = 1,
and a load resistance RL = 12 Ω, determine the power in the modulated carrier and each
significant side frequency, and sketch the power spectrum for the modulated output wave.
GIVEN:
Vc = 12 Vp
m =1
RL = 12 Ω
SOLN:
m =1
Amplitudes:
Jo
= 12 (0.77)
J1
= 12 (0.44)
J2
= 12 (0.11)
J3
= 12 (0.02)
= 9.24
= 5.28
= 1.32
= 0.24
V
V
V
V
Power Spectrum:
= (9.24)2 / 24
= (5.28)2 / 24
= (1.32)2 / 24
= (0.24)2 / 24
= 3.5574 W
= 1.1616 W
= 0.0726 W
= 0.0024 W
3.5574 W
1.1616 W
0.0726 W
0.0024 W
Power:
Po
P1
P2
P3
1.1616 W
0.0726 W
0.0024 W
31. For an angle-modulated carrier vc = 4 cos(2π300 MHz t) with 75 kHz of frequency deviation due
to the modulating signal and a single-frequency interfering signal vn = 0.2 cos(2π300.015MHz t),
determine
(a) Frequency of the demodulated interference signal.
(b) Peak and rms phase and frequency deviation due to the interfering signal.
(c) S/N ratio at the output of the FM demodulator.
a)
Ec
fc
∆f
En
fn
=4V
= 300 MHz
= 75 kHz
= 0.2 V
= 300.015 MHz
f
= fn – fc
= 300.015 – 300
= 15 kHz
f
b)
∆fpeak
∆fpeak
= f (En / Ec)
= 15 kHz (0.2 / 4)
= 7.5 kHz
∆φpeak = Vn / Vc
= 0.2 / 4
∆φpeak = 0.05 rad
RMS:
∆frms
∆frms
∆φrms
∆φrms
c)
= ∆fpeak / √2
= 7.5 kHz / √2
= 5.3 kHz
= ∆φpeak / √2
= 0.05 / √2
= 0.035 rad
S/N due to interfering tone:
S/N = Ec / En
= 4 / 0.2
= 20
S/N after demodulation:
S/N = ∆fsig /∆fnoise
= 75 / 7.5
= 10
Voltage S/N improvement:
20 / 10
=2
dB = 20 log 2
= 10 dB